Exercise 11.4 Practice
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Overview
This page provides comprehensive Ch 11: Surface Areas and Volumes - Exercise 11.4 Practice. Free NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas And Volumes Exercise 11-4. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
Volume of a Sphere and Hemisphere
Q1: Volume of Sphere
Find the volume of a sphere whose radius is:
(i) $21 \text{ cm}$
(ii) $3.5 \text{ cm}$
(i) $21 \text{ cm}$
(ii) $3.5 \text{ cm}$
Formula: Volume of Sphere $V = \frac{4}{3}\pi r^3$.
(i) Radius $r = 21 \text{ cm}$:
$V = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21$
$= 4 \times 22 \times 21 \times 21 = 88 \times 441$
$= 38808 \text{ cm}^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21$
$= 4 \times 22 \times 21 \times 21 = 88 \times 441$
$= 38808 \text{ cm}^3$.
(ii) Radius $r = 3.5 \text{ cm}$:
$V = \frac{4}{3} \times \frac{22}{7} \times (3.5)^3$
$= \frac{4}{3} \times \frac{22}{7} \times 42.875$
$= \frac{88}{21} \times 42.875 \approx 179.67 \text{ cm}^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times (3.5)^3$
$= \frac{4}{3} \times \frac{22}{7} \times 42.875$
$= \frac{88}{21} \times 42.875 \approx 179.67 \text{ cm}^3$.
(i) $38808 \text{ cm}^3$, (ii) $179.67 \text{ cm}^3$
Q2: Water Displaced
Find the amount of water displaced by a solid spherical ball of diameter:
(i) $14 \text{ cm}$
(ii) $0.42 \text{ m}$
(i) $14 \text{ cm}$
(ii) $0.42 \text{ m}$
Amount of water displaced = Volume of the sphere.
(i) Diameter $d = 14 \text{ cm} \Rightarrow r = 7 \text{ cm}$:
$V = \frac{4}{3} \times \frac{22}{7} \times 7^3$
$= \frac{4}{3} \times 22 \times 49 = \frac{4312}{3} \approx 1437.33 \text{ cm}^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times 7^3$
$= \frac{4}{3} \times 22 \times 49 = \frac{4312}{3} \approx 1437.33 \text{ cm}^3$.
(ii) Diameter $d = 0.42 \text{ m} \Rightarrow r = 0.21 \text{ m}$:
$V = \frac{4}{3} \times \frac{22}{7} \times (0.21)^3$
$= \frac{4}{3} \times \frac{22}{7} \times 0.009261$
$= 0.038808 \text{ m}^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times (0.21)^3$
$= \frac{4}{3} \times \frac{22}{7} \times 0.009261$
$= 0.038808 \text{ m}^3$.
(i) $1437.33 \text{ cm}^3$, (ii) $0.038808 \text{ m}^3$
Q3: Mass of Ball
The diameter of a metallic ball is $8.4 \text{ cm}$. What is the mass of the ball, if the density of the
metal is $5 \text{ g per cm}^3$?
Radius $r = \frac{8.4}{2} = 4.2 \text{ cm}$.
Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (4.2)^3$
$= \frac{88}{21} \times 74.088 = 310.464 \text{ cm}^3$.
Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (4.2)^3$
$= \frac{88}{21} \times 74.088 = 310.464 \text{ cm}^3$.
Mass = Volume $\times$ Density
$= 310.464 \times 5 = 1552.32 \text{ g}$.
$= 310.464 \times 5 = 1552.32 \text{ g}$.
Mass $= 1.55 \text{ kg}$ (approx)
Q4: Volume Ratio
The diameter of planet A is approximately half the diameter of planet B. What is the ratio of their
volumes?
Let diameter of planet B be $d$. Radius $R = d/2$.
Diameter of planet A $= d/2$. Radius $r = d/4$.
Diameter of planet A $= d/2$. Radius $r = d/4$.
Ratio of Volumes $= \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = \left(\frac{r}{R}\right)^3$.
$\frac{r}{R} = \frac{d/4}{d/2} = \frac{1}{2}$.
Ratio $= \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Ratio $= \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Ratio $= 1:8$
Q5: Capacity of Bowl
How many litres of milk can a hemispherical bowl of diameter $21 \text{ cm}$ hold?
Radius $r = \frac{21}{2} = 10.5 \text{ cm}$.
Volume of Hemisphere $= \frac{2}{3}\pi r^3$
$= \frac{2}{3} \times \frac{22}{7} \times (10.5)^3$
$= \frac{44}{21} \times 1157.625 = 2425.5 \text{ cm}^3$.
$= \frac{2}{3} \times \frac{22}{7} \times (10.5)^3$
$= \frac{44}{21} \times 1157.625 = 2425.5 \text{ cm}^3$.
Capacity in litres $= \frac{2425.5}{1000} = 2.4255 \text{ litres}$.
Capacity $\approx 2.43 \text{ litres}$
Q6: Volume of Material
A hemispherical tank is made up of an iron sheet $2 \text{ cm}$ thick. If the inner radius is $2 \text{
m}$, then find the volume of the iron used to make the tank.
Inner radius $r = 2 \text{ m}$.
Thickness $= 2 \text{ cm} = 0.02 \text{ m}$.
Outer radius $R = 2 + 0.02 = 2.02 \text{ m}$.
Thickness $= 2 \text{ cm} = 0.02 \text{ m}$.
Outer radius $R = 2 + 0.02 = 2.02 \text{ m}$.
Volume of iron $= \text{Outer Volume} - \text{Inner Volume}$
$= \frac{2}{3}\pi (R^3 - r^3)$
$= \frac{2}{3} \times \frac{22}{7} \times ((2.02)^3 - 2^3)$
$= \frac{44}{21} \times (8.242408 - 8)$
$= \frac{44}{21} \times 0.242408 \approx 0.5079 \text{ m}^3$.
$= \frac{2}{3}\pi (R^3 - r^3)$
$= \frac{2}{3} \times \frac{22}{7} \times ((2.02)^3 - 2^3)$
$= \frac{44}{21} \times (8.242408 - 8)$
$= \frac{44}{21} \times 0.242408 \approx 0.5079 \text{ m}^3$.
Volume $\approx 0.508 \text{ m}^3$
Q7: Volume from Surface Area
Find the volume of a sphere whose surface area is $616 \text{ cm}^2$.
Surface Area $S = 4\pi r^2 = 616$.
$4 \times \frac{22}{7} \times r^2 = 616$.
$r^2 = \frac{616 \times 7}{88} = 49 \Rightarrow r = 7 \text{ cm}$.
$4 \times \frac{22}{7} \times r^2 = 616$.
$r^2 = \frac{616 \times 7}{88} = 49 \Rightarrow r = 7 \text{ cm}$.
Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 7^3$
$= \frac{4}{3} \times 22 \times 49 = 1437.33 \text{ cm}^3$.
$= \frac{4}{3} \times 22 \times 49 = 1437.33 \text{ cm}^3$.
Volume $= 1437.33 \text{ cm}^3$
Q8: Dome Whitewashing
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of
₹$6160$. If the cost of white-washing is ₹$20$ per square metre, find the:
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
(i) Inside Surface Area:
Area $= \frac{\text{Total Cost}}{\text{Rate}} = \frac{6160}{20} = 308 \text{ m}^2$.
Area $= \frac{\text{Total Cost}}{\text{Rate}} = \frac{6160}{20} = 308 \text{ m}^2$.
(ii) Volume of air:
CSA $= 2\pi r^2 = 308$.
$2 \times \frac{22}{7} \times r^2 = 308 \Rightarrow r^2 = \frac{308 \times 7}{44} = 49$.
$r = 7 \text{ m}$.
Volume $= \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 343$
$= \frac{44}{3} \times 49 = 718.67 \text{ m}^3$.
CSA $= 2\pi r^2 = 308$.
$2 \times \frac{22}{7} \times r^2 = 308 \Rightarrow r^2 = \frac{308 \times 7}{44} = 49$.
$r = 7 \text{ m}$.
Volume $= \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 343$
$= \frac{44}{3} \times 49 = 718.67 \text{ m}^3$.
Area $= 308 \text{ m}^2$, Volume $= 718.67 \text{ m}^3$
Q9: Melting Spheres
Eight solid iron spheres, each of radius $r$ and surface area $S$ are melted to form a sphere with
surface area $S'$. Find the:
(i) radius $r'$ of the new sphere,
(ii) ratio of $S$ and $S'$.
(i) radius $r'$ of the new sphere,
(ii) ratio of $S$ and $S'$.
(i) Radius $r'$:
Volume of 8 spheres = Volume of new sphere.
$8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (r')^3$
$8r^3 = (r')^3 \Rightarrow r' = \sqrt[3]{8r^3} = 2r$.
Volume of 8 spheres = Volume of new sphere.
$8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (r')^3$
$8r^3 = (r')^3 \Rightarrow r' = \sqrt[3]{8r^3} = 2r$.
(ii) Ratio of $S$ and $S'$:
$S = 4\pi r^2$ and $S' = 4\pi (r')^2 = 4\pi (2r)^2 = 16\pi r^2$.
Ratio $\frac{S}{S'} = \frac{4\pi r^2}{16\pi r^2} = \frac{1}{4}$.
$S = 4\pi r^2$ and $S' = 4\pi (r')^2 = 4\pi (2r)^2 = 16\pi r^2$.
Ratio $\frac{S}{S'} = \frac{4\pi r^2}{16\pi r^2} = \frac{1}{4}$.
(i) $r' = 2r$, (ii) Ratio $1:4$